A    NEW    SYSTEM 

OF 

HAND-RAILING. 

CUT  SQUARE  TO  THE  PLANK,  WITHOUT  THE 
AID  OF  FALLING  MOULDS, 

A  NEW  AND  EASY  METHOD  OF  FORMING  HAND-RAILS. 

Containing  a  Large  Number  of  Illustrations  of  Hand-Kails, 
with  Full  Instructions  for  Working  Them. 


AN  OLD  STAIR-BUILDER 


REVISED     AND     CORRECTED     EDITION 


NEW    YORK: 
THE    INDUSTRIAL    PUBLICATION    COMPANY. 

1885. 


*    * 


CVUFORN 
i  A  UAUBARA 


PREFACE. 


THE  following  method  of  getting  out  wreaths  for  hand-railing,  while 
i! »!  ii  -w.  is  nut  generallyknowu  in  this  country,  which  is  to  be  regretted. 
i;\  tlii.-  method  much  time  mid  material  arc  saved,  and  the  wreaths  formed 
with  mon;  accuracy  than  cutting  them  out  by  the  older  methods,  as  any 
wreath  whatever  may  be  cut  out  square  Iroin  a  plank  of  the  same  thickness 
as  the  diameter  of  a  circle,  described  round  a  section  of  the  mil. 

In  the  annexed  cut,  A  B  shows  the  thickness  of  the  plank.  But  it  is 
1 .  1 1  •  •!•  for  the  learner  to  allow  the  plank  4  inch  thicker,  as  it  leaves  more  for 


squaring  the  rail.  By  this  method  the  same  bevel  applies  to  both  ends  of 
the  wreath  an  the  square  joint,  for  stairs  with  any  number  of  winders  in  the 
well:  tills  ensures  the  rall  rising  equally  all  round  the  weU,  and  not  quick 
in  one  part  and  flat  in  another. 

All  wreaths  are  cut  out  square  from  the  plank  with  square  joints,  and 
Boaoourate  are  the  bevels  and  joints  that  the  rail  may  he  H,,med  up  ready 
lor  polishing  before  it  leaves  the  bench. 

Seven  complete  examples  are  RivRD,  with  ample  illustrations  and 
.le,eri|,livetext .;  these  examples  IHn-  quj(..  slim,.j..n)  1o  ,,nnhjc  tho  Rt.He,u 
to  build  a  rail  over  any  kind  of  a  stair,  no  matter  l,,,xv  many  winder*  H.e.v 
may  l>e  or  how  they  are  arranged. 

New  York,  1885. 


INTRODUCTION. 


BEFOBE  proceeding  to  describe  the  method  of  obtaining  the  lines 
necessary  for  the  correct  formation  of  Hand-mils,  it  is  deemed  advis;ible 
to  present  for  the  student's  use  a  rudimentary  treatise  on  what  is  known 
as  CAKPENTEK'S  GEOMETKY,  as  it  is  felt  that  a  knowledge  of  this  will  ma- 
terially aid  the  young  beginner  to  a  better  understanding  of  the  principles 
involved  in  the  work. 

The  following  treatise,  which  is  taken  from  "Practical  Carpentry,"  is 
simple  and  easily  understood.  Let  me  add,  however,  that,  although  a 
knowledge  of  geometry  will  greatly  help  the  student  to  understand  the 
method  herein  described,  such  knowledge  is  not  absolutely  necessary,  as 
the  Art  of  Hand-railing,  as  here  presented,  may  be  efficiently  acquired 
without  it. 


TABLE   OF   CONTENTS. 


PART  I. 

PAG!? 

GEOMETRY.— Straight  Lines.— Curved  Lines.— Solids.— Compound  Lines. 
—  Parallel  Lines. — Oblique  or  Converging  Lines.— Plain'  l-'iirmv<. 
Angles.— Right  Angles.— Acute  Angles.— Obtuse  Angles.— Right-angled 
Triangles.  —  Quadrilateral  Figures.  —  Parallelograms.  —  Rectan  _ 
Squares. —  Rhomboids.—  Trapeziums.—  Trapezoids. —  Diagonals. —  Poly- 
gons. —  Pentagons.—  Hexagons.  —  Heptagons.  —  Octagons.  —  Circles.— 
Chords.  -Tangents.— Sectors.— Quadrants.— Arcs.— Concentric  and  Ec- 
centric Circles.— Altitudes.— Problems  I.  to  XXIX.  — Drawing  of  Angles, 
('(instruction  of  (ieometrical  Figures.— Bisection  of  Lines.-  Trisection 
of  Lines  and  Angles.  — Division  of  Lines  into  any  Number  of  Parts.— 
Con>t  ruction  of  Triangles,  Squares  and  Parallelograms.— Construction  of 
Proportionate  Squares.— Construction  of  Polygons.— Areas  of  Polygons. 
—Areas  of  Concentric  Rings  and  Circles.  Segments  of  Circles. —The  u-e 
of  Ordinates  for  Obtaining  Arcs  ot  Circles.  -Drawing  an  Ellipse  with  a 
Trammel.— Drawing  an  Ellipse  by  means  ot  a  String.  Same  by  Ordin- 
ates.—Raking  Ellipses.— Ovals. —Sixty-two  Illustrations.— Tangents  and 
Diagrams  for  Handrailing.  -Three  Illustrations, 7-34 


PART  II. 

Lines  and  Method  for  Making  the  Wreaths  for  Stairs  with  Four  Winders 
in  the  Well.  Four  Illustrations  and  Diagrams.— Lines  and  Method  for 
Making  the  Wreaths  for  Stairs  with  Six  Winders  in  the  Weil.  Three 
Illustrations  and  Working  Diagrams.  Lines  and  Methoi  for  Making  the 
Wivatlis  lor  Stairs  with  Three  Winders  in  Half  the  Well,  and  Landing 
in  the  Centre  of  the  Well.  Four  Illustrations  and  Working  Diagrams.— 
Method  of  Making  the  Wreaths  for  Stairs  with  One  Riser  iu  the  Centre  of 
Well.  With  Three  Illustrations  and  Working  Diagrams.— How  to  Make 
the  Wreaths  for  Level  Landing  Stairs,  the  distance  between  Centre  of 
Rails  across  the  Well  being  equal  to  the  Tread  of  a  Step,  or  nearly  so. 
With  Three  Illustrations  and  Working  Diagrams.— How  to  Make  the 
Wreath  tor  Quarter-landing  Stairs,  having  One  Winder  in  the  Well. 
Three  Illustrations  and  Working  Diagrams.  How  to  Make  a  Wreath  for 
Quarter-landing  stairs,  with  Itisers  on  the  Line  of  Spring.  Three  Illus- 
trations and  Working  Diagrams.  -  -  :i.')-65 


HAND-RAILING. 


PART  I.-GEOMETRT. 

1EFORE  a  knowledge  of  geometry  can  be  acquired,  it  will 
be  necessary  to  become  acquainted  with  some  of  the 
terms  and  definitions  used  in  the  science  of  geometry, 
and  to  this  end  the  following  terms  and  explanations  are  given, 
though  it  must  be  understood  that  these  are  only  a  few  of  the 
terms  used  in  the  science,  but  they  are  sufficient  for  our  purposes : 

1.  A  'Point  has  position  but  not  magnitude.     Practically,  it  is 
represented  by  the  smallest  visible  mark  or  dot,  but  geometrically 
understood,  it  occupies  no  space.     The  extremities  or  ends  of  lines 
are  points;    and   when  two  or  more  lines  cross  one  another,  the 
places  that  mark  their  intersections  are  also  points. 

2.  A  Line  has  length,  without  breadth  or  thickness,  and,  conse- 
quently, a  true  geometrical  line  cannot  be  exhibited ;  for  however 
finely  a  line  may  be  drawn,  it  will  always  occupy  a  certain  extent 
of  space. 

3.  A  Superficies  or  Surface  has  length  and  breadth,  but  no  thick- 
ness.    For  instance,  a  shadow  gives  a  very  good  representation  of 
a  superficies :  its  length  and  breadth  can  be  measured ;  but  it  has 
no  depth  or  substance.     The  quantity  of  space  contained  in  any 
plane  surface  is  called  its  area. 

4.  A  Plane  Superficies  is  a  flat  surface,  which  will  coincide  with 
a  straight  line  in  every  direction. 


10  HAND-RAILING. 

5.  A  Curved  Superficies  is  an  uneven  surface,  or  such  as  will  not 
coincide  with  a  straight  line  in  all  directions.     By  the  term  surface 
is  generally  understood  the  outside  of  any  body  or  object;  as,  for 
instance,  the  exterior  of  a  brick  or  stone,  the  boundaries  of  which 
are  represented  by  lines,  either  straight  or  curved,  according  to  the 
form  of  the  object.     We  must  always  bear  in  mind,  however,  that 
the  lines  thus  bounding  the  figure  occupy  no  part  of  the  surface  ; 
hence   the   lines   or   points    traced    or   marked  on    any  body    or 
surface,   are    merely   symbols    of    the   true    geometrical  lines   or 
points. 

6.  A  Solid  is  anything  which  has  length,  breadth  and  thickness ; 
consequently,  the  term   may  be  applied  to  any  visible  object  con- 
taining substance ;  but,  practically,  it  is  understood  to  signify  the 
solid  contents  or  measurement  contained  within  the  different  sur- 
faces of  which  any  body  is  formed. 

7.  Lines  may  be  drawn  in  any  direction,  and  are  termed  straight, 
curved,  mixed,  concave,  or  convex  lines,  according  as  they  corres- 
pond to  the  following  definitions. 

8.  A  Straight  Line  is  one  every  part  of  which  A —  — B 
lies  in  the  same  direction  between   its  extremities,           Flg-  I- 
and  is,  of  course,  the  shortest  distance   between  two  pbints,  as 
from  A  to  B,  Fig.  i. 

9.  A  Curved  Line  is  such  that  it  does  not  lie  in  a  straight  direc- 
tion between  its  extremities,  but  is  continually  changing  by  inflec- 
tion.    It  may  be  either  regular  or  irregular. 

10.  A  Mixed  or  Compound  Line  is  composed  of  straight  and 
curved  lines,  connected  in  any  form. 

11.  A  Concave  or  Convex  Line  is  such  that  it  cannot  be  cut  by 
a  straight  line  in  more  than  two  points ;    the  concave  or  hollow 
side  is  turned  towards  the  straight  line,  while  the  convex  or  swell- 
ing side  looks  away  from  it.     For  instance,  the  inside  of  a  basin  is 
concave — the  outside  of  a  ball  is  convex. 

12.  Parallel  Straight  Lines  have  no  inclination,  but  are  every- 
where at  an  equal  distance  from  each  other;  consequently  they 
can  never  meet,  though  produced  or  continued  to  infinity  in  either 
or  both  directions.     Parallel  lines  may  be  either  straight  or  curved, 


HAND-RAILING.  II 

provided  they  are  equally  distant  from  each  other  throughout  their 
extension. 

13.  Oblique  or  Converging  Lines  are  straight  lines,  which,  if  con- 
tinued, being  in  the  same  plane,  change  their  distance  so  as  to 
meet  or  intersect  each  other. 

14.  A  Plane  Figure,  Scheme,  or  Diagram,  is  the  lineal  representa- 
tion of  any  object  on  a  plane  surface.     If  it  is  bounded  by  straight 
lines,  it  is  called  a  rectilineal  figure ;  and  if  by  curved  lines,  a 
curvilineal  figure. 

15.  An  Angle  is  formed  by  the  inclination  of  two  lines  meeting 
in  a  point :  the  lines  thus  forming  the  angle  are  called  the  sides ; 
and  the  point  where  the  lines  meet  is  called  the  vertex  or  angular 
point. 

When  an  angle  is  expressed  by  three  letters,  as  A  u  c,  Fig.  2,  the 
middle  letter  u  should  always  denote  the  angular  point:    where 


Fig.  3. 


Fig.  4. 


Fig.  5. 


there  is  only  one  angle,  it  may  be  expressed  more  concisely  by  a 
letter  placed  at  the  angular  point  only,  as  the  angle  at  A,  Fig.  3. 

1 6.  The  quantity  of  an  angle  is  estimated  by  the  arc  of  any 
circle  contained  between  the  two  sides  or  lines  forming  the  angle ; 
the  junction  of  the  two  lines,  or  vertex  of  the  angle,  being  the 
centre  from  which  the  arc  is  described.  As  the  circumferences  of 
all  circles  are  proportional  to  their  diameters,  the  arcs  of  similar 
sectors  also  bear  the  same  proportion  to  their  respective  circum- 
ferences; and,  consequently,  are  proportional  to  their  diameters, 
and,  of  course,  also  to  their  radii  or  semi-diameters.  Hence,  the 


12  HAND-RAILING. 

proportion  which  the  arc  of  any  circle  bears  to  the  circumference 
of  that  circle,  determines  the  magnitude  of  the  angle.  From  this 
it  is  evident  that  the  quantity  or  magnitude  of  angles  does  not  de- 
pend upon  the  length  of  the  sides  or  radii  forming  them,  but 
wholly  upon  the  number  of  degrees  contained  in  the  arc  cut  from 
the  circumference  of  the  circle  by  the  opening  of  these  lines.  The 
circumference  of  every  circle  is  divided  by  mathematicians  into 
360  equal  parts,  called  degrees ;  each  degree  being  again  subdi- 
vided into  60  equal  parts,  called  minutes,  and  each  minute  into  60 
parts,  called  seconds.  Hence,  it  follows  that  the  arc  of  a  quarter 
circle  or  quadrant  includes  90  degrees ;  that  is,  one-fourth  part  of 
360  degrees.  By  dividing  a  quarter  circle;  that  is,  the  portion  of 
the  circumference  of  any  circle  contained  between  two  radii  form- 
ing a  right  angle,  into  90  equal  parts,  or,  as  is  shown  in  Fig.  4,  into 
nine  equal  parts  of  10  degrees  each,  then  drawing  straight  lines 
from  the  centre  through  each  point  of  division  in  the  arc ;  the  right 
angle  will  be  divided  into  nine  equal  angles,  each  containing  10 
degrees.  Thus,  suppose  B  c  the  horizontal  line,  and  A  n  the  per- 
pendicular ascending  from  it,  any  line  drawn  from  B — the  centre 
from  which  the  arc  is  described — to  any  point  in  its  circumference, 
determines  the  degree  of  inclination  or  angle  formed  between  it 
and  the  horizontal  line  B  c.  Thus,  a  line  from  the  centre  B  to  the 
tenth  degree,  separates  an  angle  of  10  degrees,  and  so  on.  In  this 
manner  the  various  slopes  or  inclinations  of  angles  are  defined. 

17.  A  Right  Angle  is  produced  by  one  straight  line   standing 
upon  another,  so  as  to  make  the  adjacent   angles   equal.     This  is 
what   workmen   call   "  square,"  and  is  the  most  useful  figure  they 
employ. 

1 8.  An  Acute  Angle   is  less  than  a  right  angle,  or  less  than  90 
degrees. 

19.  An  Obtuse  Angle  is  greater  than  a  right  angle  or  square,  oi 
more  than  90  degrees. 

The  number  of  degrees  by  which  an  angle  is  less  than  90  de- 
grees is  called  the  complement  of  the  angle.  Also,  the  difference 
between  an  obtuse  angle  and  a  semicircle,  or  180  degrees,  is  called 
the  supplement  of  that  angle. 


13 

20.  Plane  Figures  are  bounded  by  straight  lines,  and  are  named 
according  to  the  number  of  sides  which   they  contain.     Thus,  the 
space   included   within  three    straight  lines,  forming  three  angles, 
is  called  a  trilateral  figure  or  triangle. 

21.  A  Right-Angled  Triangle   has   one  right   angle:    the   sides 
forming  the  right  angle  are  called  the  base  and  perpendicular;  and 
the  side  opposite  the   right  angle  is  named  the    hypothenuse.     An 
equilateral    triangle  has  all  its  sides  of  equal  length.     An  isosceles 
triangle  has   only  two   sides   equal ;  a  scalene  triangle  has  all  its 
sides  unequal.     An   acute-angled  triangle  has  all  its  angles  acute, 
and    an    obtused-angled     triangle    has    one    of     its    angles    only 
obtuse.  » 

The  triangle  is  one  of  the  most  useful  geometrical  figures  for  the 
mechanic  in  taking  dimensions;  for  since  all  figures  that  are 
bounded  by  straight  lines  are  capable  of  being  divided  into  tri- 
angles, and  as  the  form  of  a  triangle  cannot  be  altered  without 
changing  the  length  of  some  one  of  its  sides,  it  follows  that  the  true 
form  of  any  figure  can  be  preserved  if  the  length  of  the  s:des  of 
the  different  triangles  into  which  it  is  divided  is  kaown ;  and  the 
area  of  any  triangle  can  easily  be  ascertained  by  the  same  rr.le,  as 
will  be  shown  further  on. 

Quadrilateral  Figures  are  literally  four-sided  figures.  They  are 
also  called  quadrangles,  because  they  have  four  angles. 

22.  A  Parallelogram  is  a  figure  whose  opposite  sides  are  parallel, 
as  A  B  c  D,  Fig.  5. 

23.  A  Rectangle  is  a  parallelogram  having  four  right  angles,  as  A 
BCD,  in  Fig.  5. 

24.  A   Square    is  an   equilateral   rectangle,  having  all  its  sidi-s 
equal,  like  Fig.  5. 

25.  An  Oblong  is  a  rectangle  whose  adjacent  sides  are  unequal, 
as  the  parallelogram  shown  at  Fig.  TO. 

26.  A  Rhombus  is  an  oblique-angled  figure,    or   parallelogram 
having  four  equal  sides,  whose  opposite  angles  only  are  equal,  as 
c,  Fig.  6. 

27.  A  Rhomboid  is  an  oblique-angled  parallelogram,  of  which  the 
adjoining  sides  are  unequal,  as  D,  Fig.  7. 


HAND-RAILING. 


.28.  A  Trapezium  is  an  irregular  quadrilateral  figure,  having  no 
two  sides  parallel,  as  E,  Fig.  8. 

29.  A  Trapczoid  is  a  quadrilateral  figure,  which  has  two  of  its 
opposite  sides  parallel,  and  the  remaining  two  neither  parallel  nor 
equal  to  one  another,  as  F,  Fig.  9. 


Fig.  7. 


Fig.  8. 


30.  A  Diagonal  is  a  straight   line   drawn   between  two  opposite 
angular  points  of  a  quadrilateral  figure,  or  between  any  two  angular 
points  of  -a  polygon.     Should   the  figure  be  a  parallelogram,  the 
diagonal  will  divide  it  into  two  equal  triangles,  the  opposite  sides 
and   angles  of  which  will  be  equal  to  one   another.     Let  A  B  c  D, 
Fig.  10,  be  a  parallelogram;  join  A  c,  then  A  c  is  a  diagonal,  and 
the  triangles  A  D  c,  A  B  c,  into  which  it  divides  the  parallelogram, 
are  equal. 

31.  A  plane  figure,  bounded  by  more  than  four  straight  lines,  is 
called  a  Polygon.     A  regular   polygon    has  all  its  sides  equal,  and 
consequently  its   angles   are   also  equal,  as  K,  L,  M,  and  N,  Figs. 


Fig.  11. 


Fig.  12. 


Fig.  13. 


12-15.  An  irregular  polygon  has  its  sides  and  angles  unequal,  as 
H,  Fig.  1 1.  Polygons  are  named  according  to  the  number  of  their 
sides  or  angles,  as  follows  : — 

32.  A  Pentagon  is  a  polygon  of  five  sides,  as  H  or  K,  Figs,  u,  12. 

33.  A  Hexagon  is  a  polygon  of  six  sides,  as  L,  Fig.  13. 

34.  A  Heptagon  has  seven  sides,  as  M,  Fig.  14. 


HAND-RAILING. 


35.  An  Octagon  has  eight  sides,  as  N,  Fig.  15. 

An  Enneagnt  has  nine,  a  Decagon  ten,  an  Undecagon  eleven,  and 
a  Dodecagon  twelve  sides.  Figures  having  more  than  twelve  sides 
are  generally  designated  Polygons,  or  many-angled  figures. 

36.  A  Circle  is  a  plane  figure  bounded  by  one  uniformly  curv    I 
line,  bed  (Fig.  16),  called  the  circumference,  every  part  of  which 
is  equally  distant  from  a  point  "within  it,  called  the  centre,  as  a. 

37.  The  Radius  of  a  circle  is  a  straight  line  drawn    from   the 
centre  to  the  circumference ;  hence,  all  the  radii  (plural  for  radius) 
of  the  same  circle  are  equal,  as  b  a,  c  a,  e  a,f  a,  in  Fig.  16. 

38.  The  Diameter  of  a  circle  is  a  straight  line  drawn  through  the 
centre,  and  terminated  on  each  side  by  the  circumference;  conse- 

X 

4 


ig.  16. 


Fig.  }7f 


quently  the  diameter  is  exactly  twice  the  length  of  the  radius  ;  and 
hence  the  radius  is  sometimes  called  the  semi-diameter.  (See  b  a  c^ 
Fig.  1 6.) 

49.  The  Chord  or  Subtens  of  an  arc  is  any  straight  line  drawn 
from  one  point  in  the  circumference  of  a  circle  to  another,  joining 
the  extremities  of  the  arc,  and  dividing  the  circle  either  into  two 
equal,  or  two  unequal  parts.  If  into  equal  parts,  the  chord  is  also 
the  diameter,  and  the  space  included  between  the  arc  and  the  di- 
ameter, on  either  side  of  it,  is  called  a  semicircle,  as  bae'm  Fig.  16. 
If  the  parts  cut  off  by  the  chord  are  unequal,  each  of  them  is  called 
a  segment  of  the  circle.  The  same  chord  is  therefore  common  to 
two  arcs  and  two  segments ;  but,  unless  when  stated  otherwise,  it 
is  always  understood  that  the  lesser  arc  or  segment  is  spoken  of,  as 
in  Fig.  1 6,  the  chord  c  d  is  the  chord  of  the  arc  c  e  d. 

If  a  straight  line  be  drawn  from  the  centre  of  a  circle  to  meet 
the  chord  of  an  arc  perpendicularly,  as  afy  in  Fig.  16,  it  will  divide 
the  chord  into  two  equal  parts,  and  if  the  straight  line  be  produced 


1 6  (  HAND- RAILING. 

to  meet   the  arc,  it  will   also    divide  it  into   two  equal   parts,  as 
cfjd. 

Each  half  of  the  chord  is  catted  the  sine  of  the  half-arc  to  which 
it  is  opposite;  and  the  line  drawn  from  the  centre  to  meet  the 
chord  perpendicularly,  is  called  the  co-sine  of  the  half-arc.  Con- 
sequently, the  radius,  the  sine,  and  co-sine  of  an  arc  form  a  right 
angle. 

40.  Any  line  which    cuts  the  circumference  in  two  points,  or  a 
chord   lengthened  out  so  as  to  extend    beyond    the  boundaries  of 
the  circle,  such  as  g  h  in  Fig.  17,  is  sometimes  called  a  Secant. 
But,  in  trigonometry,  the   secant  is  a  line  drawn  from  the  centre 
through  one  extremity  of  the  arc,  so  as  to  meet  the  tangent  which  is 
drawn  from   the   other   extremity  at  right   angles  to   the   radius. 
Thus,  vcb  is  the  secant  of  the  arc  c  e,  or  the  angle  c  $  e,  in  Fig.  17. 

41.  A  Tangent  is  any  straight  line  which  touches  the  circumfer- 
ence of  a  circle  in  one  point,  which  is  called  the  point  of  contact, 
as  in  the  tangent  line  e  b,  Fig.  17. 

42.  A  Sector  is  the  space  included  between  any  two  radii,  and 
that  portion  of  the  circumference  comprised  between  them :  c  e  v' 
is  a  sector  of  the  circle  afc  e,  Fig.  17. 

43.  A  Quadrant,  or  quarter  of  a  circle,  is  a  sector  bounded  by 
two  radii,  forming  a  right   angle  at  the  centre,  and   having   one- 
fourth  part  of  the  circumference  for  its  arc,  as  F/V,  Fig.    7. 

44.  An  Arc,  or  Arch,  is  any  portion  of  the  circumference  of  a 
circle,  as  c  d  e,  Fig.  17. 

It  may  not  be  improper  to  remark  here  that  the  terms  circle  and 
circumference  are  frequently  misapplied.  Thus  we  say,  describe  a 
circle  from  a  given  point,  etc.,  instead  of  saying  describe  the  cir- 
cumference of  a  circle — the  circumference  being  the  curved  line 
thus  described,  everywhere  equally  distant  from  a  point  within  it, 
called  the  centre ;  whereas  the  circle  is  properly  the  superficial 
space  included  within  that  circumference. 

45.  Concentric  Circles  are  circles   within  circles,  described  from 
the  same  centre;  consequently,  their  circumferences  are  parallel  to 
one  another,  as  Fig.  18. 

46.  Eccentric  Circles  are  those  which  are  not  described  from  the 


HAND-FAILING.  iy 

same  centre;  any  point  which  is  not  the  centre  is  also  eccentric  in 
reference  to  the  circumference  of  that  circle.  Eccentric  circles 
may  also  be  tangent  circles ;  that  is,  such  as  come  in  contact  in 
one  point  only,  as  Fig.  19. 

47.  Altitude.  The  height  of  a  triangle  or  other  figure  is  called 
its  altitude.  To  measure  the  altitude,  let  fall  a  straight  line  from 
the  vertex,  or  highest  point  in  the  figure,  perpendicular  to  the 
base  or  opposite  side ;  or  to  the  base  continued,  as  at  B  D,  Fig.  20, 
should  the  form  of  the  figure  require  its  extension.  Thus  c  D  is  the 
altitude  of  the  triangle  ABC. 

We  have  now  described  all  the  figures  we  shall  require  for  the 
purpose  of  thoroughly  understanding  all  that  will  follow  in  this 
book ;  but  we  would  like  to  say  right  here  that  the  student  who 
has  time  should  not  stop  at  this  point  in  the  study  of  geometry,  for 
the  time  spent  in  obtaining  a  thorough  knowledge  of  this  useful 


Fig.  18.  Fig.  19.  Fig.  20. 

science  will  bring  in  better  returns  in  enjoyment  and  money,  than 
if  expended  for  any  other  purpose. 

We  will  now  proceed  to  explain  how  the  figures  we  have  de- 
scribed can  be  constructed.  There  are  several  ways  of  constructing 
nearly  every  figure  we  produce,  but  we  have  chosen  those  methods 
that  seemed  to  us  the  best,  and  to  save  space  have  given  as  few 
examples  as  possible  consistent  with  efficiency. 

PROBLEM  I. — Through  a  given  point  c  (Fig.  18  a),  to  draw  a 
straight  line  pat  allel  to  a  given  straight  line  A  B. 

In  A  B  (Fig.  1 8  a)  take  any  point  <t,  and  from  d  as  a  centre  with 
the  radius  d  c,  describe  an  arc  c  <?,  cutting  A  B  in  f,  and  from  c  as  a 


i8 


HAND-RAILING, 


centre,  with  the  same  radius,  describe  the  arc //D,  make  dT>  equal  to 
c  e,  join  c  D,  and  it  will  be  parallel  to  A  r». 

PROBLEM  II. — To  make  an  angle  equal  to  a  given  rectilineal  angle. 

From  a  given  point  E  (Fig.  19^),  upon  the  straight  line  E  K,  to 
make  an  angle  equal  to  the  given  angle  ABC.  From  the  angular 
point  B,  with  any  radius,  describe  the  arc<?/  cutting  B  c  and  B  A  in 
the  points  e  and/  From  the  point  E  on  E  F  with  the^same  radius, 


Fig.  18  a. 


Fig,  19  a. 


/C 


Fig.  20  «. 

describe  the  arc  kg,  and  make  it  equal  to  the  arc  e  f\  then  from  E, 
through  g,  draw  the  line  E  D  :  the  angle  D  E  F  will  be  equal  to  the 
angle  ABC. 

PROBLEM  III. — To  bisect  a  given  angle. 

Let  ABC  (Fig.  20  a)  be  the  given  angle.  From  the  angular  point 
B,  with  any  radius,  describe  an  arc  cutting  B  A  and  B  c  in  the 
points  //and  e;  also,  from  the  points  ^/and  e  as  centres,  with  any 
radius  greater  than  half  the  distance  between  them,  describe  arcs 
cutting  each  other  in/;  through  the  points  of  intersection/  draw 
B/D  :  the  angle  A  B  c  is  bisected  by  the  straight  line  B  D;  that  is, 
it  is  divided  into  two  equal  angles,  A  B  D  and  c  B  D. 

PROBLEM  IV. — To  trisect  or  divide  a  right  angle  into  tJiree  equal 
angles. 

Let  ABC  (Fig.  21)  be  the  given  right  angle.    From  the  angular 


HAND-RAILING. 


point  B,  with  any  radius,  describe  an  arc  cutting  B  A  and  B  c  in  the 
points  d  and  g\  from  the  points  d  a.ndg,  with  the  radius  B  d  or  B  gt 
describe  the  arcs  cutting  the  arc  d gine  and/;  join  i;  e  and  B/: 
these  lines  will  trisect  the  angle  A  B  c,  or  divide  it  into  three  equal 
angles. 

The  trisection  of  an  angle  can  be  effected  by  means  of  elementary 
geometry  only  in  a  very  few  cases;  such,  for  instance,  as  those 
where  the  arc  which  measures  the  proposed  angle  is  a  whole  circle, 
or  a  half,  a  fourth,  or  a  fifth  part  of  the  circumference.  Any  angle 
of  a  pentagon  is  trisected  by  diagonals,  drawn  to  its  opposite 
angles. 

PROBLEM  V. — From  a  given  point  c,  in  a  given  straight  line  A  B, 
to  erect  a  perpendicular. 

From  the  point  c  (Fig.  22),  with  any  radius  less  than  c  A  or  c  B, 
describe  arcs  cutting  the  given  line  A  B  in  d  and  e  •  from  these 


Fig.  21. 


Fig.  22. 


points  as  centres,  with  a  radius  greater  than  c  d  or  c  e,  describe  arcs 
intersecting  each  other  in/:  join  c/,  and  this  line  will  be  the  per- 
pendicular required. 

Another  Method. — To  draw  a  right  angle  or  erect  a  perpendicular 
by  means  of  any  scale  of  equal  parts,  or  standard  measure  of  inches, 
feet,  yards,  etc.,  by  setting  off  distances  in  prooortion  to  the  num- 
bers 3,  4  and  5,  or  6,  8  and  10,  or  any  numbers  whose  squares  cor- 
respond to  the  sides  and  hypothenuse  of  a  right-angled  triangle. 

From  any  scale  of  equal  parts,  as  that  represented  by  the  line  D 
(Fig.  23),  which  contains  5,  set  off  from  B,  on  the  line  A  B,  the  dis- 
tance B  e,  equal  to  3  of  these  parts;  then  from  B,  with  a  radius 
equal  to  4  of  the  same  parts,  describe  the  arc  a  £.;  also  from  e  as  a 


centre,  with  a  radius  equal  to  5  parts,  describe  another  arc  inter- 
secting the  former  in  c ;  lastly  join  D  c ;  the  line  u  c  will  be  per- 
pendicular to  A  B. 

This  mode  of  drawing  right  angles  is  more  troublesome  upon 
paper  than  the  method  previously  given  ;  but  in  laying  out  grounds 
or  foundations  of  buildings  it  is  often  very  useful,  since  only  with  a 
ten-foot  pole,  tape  line,  or  chain,  perpendiculars  may  be  set  out 
very  accurately.  The  method  is  demonstrated  thus  : — The  square 
of  the  hypothenuse,  or  longest  side  of  a  right-angled  triangle,  being 
equal  to  the  sum  of  the  squares  of  the  other  two  sides,  the  same 
property  must  always  be  inherent  in  any  three  numbers,  of  which 


o 

Fig.  23. 

the  squares  of  the  two  lesser  numbers,  added  together,  are  equal  to 
the  square  of  the  greater.  For  example,  take  the  numbers  3,  4, 
and  5 ;  the  square  of  3  is  9,  and  the  square  of  4  is  16 ;  16  and  9, 
added  together  make  25,  which  is  5  times  5,  or  the  square  of  the 
greater  number.  Although  these  numbers,  or  any  multiple  of 
them,  such  as  6,  8,  10,  or  12,  16,  20,  etc.,  are  the  most  simple,  and 
most  easily  retained  in  the  memory,  yet  there  are  other  numbers, 
very  different  in  proportion,  which  can  be  made  to  serve  the  same 
purpose.  Let  n  denote  any  number  ;  then  «*  +  i,  «* —  i,  and  2//, 
will  represent  the  hypothenuse,  base,  and  perpendicular  of  a  right- 
angled  triangle.  Suppose  ;/  =  6,  then  n2  +  i  =37,  n2 —  i  =35, 
and  2»=  12:  hence,  37,  35,  and  12  are  the  sides  of  a  right-angled 
triangle.  A  knowledge  of  this  problem  will  often  prove  of  the 
greatest  service  to  the  workman. 


HAND-KAI!  21 

PROBLEM  VI. —  7l>  Insut  <r  ghrn  straight  line. 

Let  A  B  (Fig.  24)  be  the  given  straight  line.  From  the  extreme 
points  A  and  B  as  centres,  with  any  equal  radii  greater  than  half 
the  length  of  A  r>,  describe  arcs  cutting  each  other  in  c  and  D:  a 
straight  line  drawn  through  the  points  of  intersection  c  and  D,  will 
bisect  the  line  A  u  in  e. 

PROBLEM  VII. — To  divide  a  given  line  into  any  number  of  equal 
parts. 

Let  A  B  (Fig.  25)  be  the  given  line  to  be  divided  into  five  equal 
parts.  From  the  point  A  draw  the  straight  line  A  c,  forming  any 
angle  with  A  B.  On  the  line  A  c,  with  any  convenient  opening  of 
the  compasses,  set  off  five  equal  parts  towards  c;  join  the  extreme 


Fig.  25.  Fig.  26.  Fig.  27 

points  c  B;  through  the  remaining  points  i,  2,  3,  and  4,  draw  lines 
parallel  to  B  c,  cutting  A  i;  in  the  corresponding  points,  i,  2,  3, 
and  4  :  A  B  will  be  divided  into  five  equal  parts,  as  required. 

There  are  several  other  methods  by  which  lines  may  be  divided 
into  equal  parts ;  they  are  not  necessary,  however,  for  our  purpose, 
so  we  will  content  ourselves  with  showing  how  this  problem  may 
be  used  for  changing  the  scales  of  drawings  whenever  such  change 
is  desired.  Let  A  B  (Fig.  26)  represent  the  length  of  one  scale  or 
drawing,  divided  into  the  given  parts  A  d,  rf  e,  ef,fg,  gh,  and  //  B ; 
and  D  E  the  length  of  another  scale  or  drawing  required  to  be  di- 
vided into  similar  parts.  From  the  point  B  draw  a  line  B  C  =  D  E, 
and  forming  any  angle  with  A  r, ;  join  A  c,  and  through  the  points 
^»  ^/j  C>  an(l  h,  draw  d  k,  el,  ftn,  g  n,  h  <?,  parallel  to  A  c ;  and 
the  parts  c  k,  k  /,  I  m,  etc.,  will  be  to  each  other,  or  to  the  whole 
line  i;  c,  as  the  lines  A  d,  d  c,  c  f,  dr..  are  to  each  other,  or  to  the 
given  line  or  scale  A  B.  By  this  method,  as  will  be  evident  from 


HAND-RAILING. 


the  figure,  similar  divisions  can  be  obtained  in  lines  of  any  given 
length. 

PROBLEM  VIII. — To  describe  an  equilateral  triangle  upon  a  given 
straight  line. 

Let  A  B  (Fig.  27)  be  the  given  straight  line.  From  the  points 
A  and  B,  with  a  radius  equal  to  A  B,  describe  arcs  intersecting  each 
other  in  the  point  c.  Join  c  A  and  c  B,  and  ABC  will  be  the 
equilateral  triangle  required. 

PROBLEM  IX. — To  construct  a  triangle  whose  sides  shall  be  equal 
to  three  given  lines,  F,  E,  D. 

Draw  A  B  (Fig.  28)  equal  to  the  given  line  F.  From  A  as  a 
centre,  with  a  radius  equal  to  the  line  E,  describe  an  are;  then 


Pig.  28. 


Fig.  29. 


from  B  as  a  centre,  with  a  radius  equal  to  the  line  D,  describe  an- 
other arc  intersecting  the  former  in  c ;  join  c  A  and  c  B,  and  ABC 
will  be  the  triangle  required. 

PROBLEM  X. — To  describe  a  rectangle  or  parallelogram  having  one 
of  its  sides  equal  to  a  given  line,  and  its  area  equal  to  that  of  a  given 
rectangle. 

Let  A  B  (Fig.  29)  be  the  given  line,  and  c  D  E  F  the  given 
rectangle.  Produce  c  E  to  G,  making  E  G  equal  to  A  B;  from  G 
draw  G  K  parallel  to  E  F,  and  meeting  D  F  produced  in  n.  Draw 
the  diagonal  G  F,  extending  it  to  meet  c  D  produced  in  L  ;  also 
draw  L  K  parallel  to  D  H,  and  produce  E  F  till  it  meet  L  K  in  M  j 
then  F  M  K  H  is  the  rectangle  required. 


HA  \D-1-:  A 1 1, 1  NT,. 


Equal  and  similar  rhomboids  or  parallelograms  of  any  dimen- 
sions may  be  drawn  after  the  same  manner,  seeing  the  comple- 
ments of  the  parallelograms  which  are  described  on  or  about  the 
diagonal  of  any  parallelogram,  are  always  equal  to  each  other; 
while  the  parallelograms  themselves  are  always  similar  to  each 
other,  and  to  the  original  parallelogram  about  the  diagonal  of 
which  they  are  constructed.  Thus,  in  the  parallelogram  c  c;  K  i, 
the  complements  c  E  F  n  and  F  M  K  H  are  always  equal,  while  the 
parallelograms  K  F  H  G  and  D  F  M  L  about  the  diagonal  o  L,  are 
always  similar  to  each  other,  and  to  the  whole  parallelogram 
c  G  K  L. 

PROBLEM   XI. — To  describe  a  square  equal  to  Iwo  given  squares. 

Let  A  and  B  (Fig.  30)  be  the  given  squares.  Place  them  so  that 
a  side  of  each  may  form  the  right  angle  D  c  E  ;  join  D  E,  and  upon 
this  hypothenuse  descnbe  the  square  D  E  G  F,  and  it  will  be  equal 
to  the  sum  of  the  squares  A  and  B,  which  are 
constructed  upon  the  legs  of  the  right-angled 
triangle  D  c  E.  In  the  same  manner,  any 
other  rectilineal  figure,  or  even  circle,  may 
be  found  equal  to  the  sum  of  other  two  simi- 
lar figures  or  circles.  Suppose  the  lines  c  D 
and  c  E  to  be  the  diameters  of  two  circles, 
then  D  F.  will  be  the  diameter  of  s.  third, 
equal  in  area  to  the  other  two  circles.  Or 
suppose  c  D  and  c  E  to  be  the  like  sides  of 
any  two  similar  figures,  then  D  E  will  be  the 
corresponding  side  of  another  similar  figure 
equal  to  both  the  former. 

PROBLEM  XII. —  To  descnbe  a  square  equal  to  any  number  of  given 
squares. 

Let  it  be  required  to  construct  a  square  equal  to  the  three  givt-n 
squares  A,  B,  and  c  (Fig.  31).  Take  the  line  n  E,  equal  to  the  side 
of  the  square  c.  From  the  extremity  D  erect  D  F  perpendicular  to 
D  E,  and  equal  to  the  side  of  the  square  B  ;  join  E  F;  then  a  square 
described  upon  this  line  will  be  equal  to  the  sum  of  the  two  given 
squares  c  and  B.  Again,  upon  the  straight  line  E  F  erect  the  per- 


Fig.  30. 


HAND-RAILING. 


Fig.  31. 


pendicular  F  G,  equal  to  the  side  of  the 
third  given  square  A  ;  and  join  G  E,  which 
will  be  the  side  of  the  square  G  E  H  K, 
equal  in  area  to  A,  B,  and  c.  Proceed  in 
the  same  way  for  any  number  of  given 
squares. 

PROBLEM  XIII. —  Upon  a  given  straiglit 
line  to  describe  a  regular  polygon. 

To  produce  a  regular  pentagon  draw 
A  B  to  c  (Fig.  32),  so  that  B  c  may  be 
equal  to  A  B  ;  from  B  as  a  centre,  with  the 
radius  B  A  or  B  c  describe  the  se'micircle 
ADC;  divide  the  semi-circumference 
ADC  into  as  many  equal  parts  as  there 
are  parts  in  the  required  polygon,  which, 
in  the  present  case,  will  be  five ;  through  the  second  division  from 
C  draw  the  straight  line  B  D,  which  will  form  another  side  of  the 
figure.  Bisect  A  B  at  e,  and  B  D  at/  and  draw  e  G  and  /G  per- 
pendicular to  A  B  and  B  D  ;  then  G,  the 
point  of  intersection,  is  the  centre  of  a 
circle,  of  which  A  B  and  D  are  points 
in  the  circumference.  From  G,  with  a 
radius  equal  to  its  distance  from  any  of 
these  points,  describe  the  circumference 
A  B  D  H  K  ;  then  producing  the  dotted 
lines  from  the  centre  B,  through  the 
remaining  divisions  in  the  semicircle 
A  D  c,  so  as  to  meet  the  circumference  of  which  G  is  the  centre, 
in  H  and  K,  these  points  will  divide  the  circle  A  B  D  H  K  into  the 
number  of  parts  required,  each  part  behip'  equal  to  the  given  side 
of  the  pentagon. 

From  the  preceding  example  it  is  evident  that  polygons  of  any 
number  of  sides  may  be  constructed  upon  the  same  principles,  be- 
cause the  circumferences  of  all  circles,  when  divided  into  the  same 
number  of  equal  parts,  produce  equal  angles;  and,  consequently, 
by  dividing  the  semi-circumference  of  any  given  circle  into  th^ 


number  of  parts  required,  t\vo  of  these  parts  will  form  an  angle 
which  will  be  subtended  by  its  corresponding  part  of  the  whole 
circumference.  And  as  all  Tegular  polygons  can  be  inscribed  in  a 
circle,  it  must  necessarily  follow,  that  if  a  circle  be  described 
through  three  given  angles  of  that  polygon,  it  will  contain  the 
number  of  sides  or  angles  required. 

The  above  is  a  general  rule,  by  which  all  regular  polygons  may 
be  described  upon  a  given  straight  line;  but  there  are  other 
methods  by  which  many  of  them  may  be  more  expeditiously  con- 
structed, as  shown  in  the  following  examples : — 

PROBLEM  XIV. —  Upon  a  given  straight  line  to  describe  a  regular 
pentagon. 

Let  A  B  (Fig.  33)  be  the  given  straight  line;  from  its  extremity 
B  erect  B  c  perpendicular  to  A  B,  and  equal 
to  its  half.  Join  A  c,  and  produce  it  till 
c  d  be  equal  to  B  c,  or  half  the  given  line 
A  B.  From  A  and  B  as  centres,  with  a 
radius  equal  to  B  d,  describe  arcs  inter- 
secting each  other  in  e,  which  will  be  the 
centre  of  the  circumscribing  circle  A  B  F 
o  H.  The  side  A  B  applied  successively 
to  this  circumference,  will  give  the  angu- 
lar  points  of  the  pentagon;  and  these 
being  connected  by  straight  lines  will  complete  the  figure. 

PROBLEM  XV. —  Upon  a  given  straight  line  to  describe  a  regular 
hexagon. 

Let  A  B  (Fig.  34)  be  the  given  straight  line.  From  the  extremi- 
ties A  and  B  as  centres,  with  the  radius  A  B  describe  arcs  cutting 
each  other  in  g.  Again  from  g,  the  point  of  intersection,  with  the 
sa-r>e  radius,  describe  the  circle  ABC,  which  will  contain  the  given 
side  A  B  six  times  when  applied  to  its  circumference,  and  will  be 
the  heA'agoiurequired. 

PROBLEM  XVI. — To  describe  a  regular  octagon  upon  a  given 
sf might  line. 

Let  A  B  (Fig.  35)  be  the  given  line.  From  the  extremities  A 
and  B  erect  the  perpendiculars  A  E  and  B  F;  extend  the  given 


2  6 


HAND-RAILING. 


line  both  ways  to  k  and  /,  forming  external  right  angles  with  the 
lines  A  E  and  B  F.  Bisect  these  external  right  angles,  making  each 
of  the  bisecting  lines  A  H  and  B  c  equal  to  the  given  line  A  B. 
Draw  H  G  and  c  D  parallel  to  A  E  or  B  F,  and  each  equal  in  length 
to  A  B.  From  G  draw  G  E  parallel  to  B  c,  and  intersecting  A  E  in 
E,  and  from  D  draw  D  F  parallel  to  A  H,  intersecting  B  F  in  F. 
Join  E  F,  and  ABCDFEGHis  the  octagon  required.  Or  from  D 


Fig  34. 


Fig.  35. 


Fig.  36. 


and  G  as  centres,  with  the  given  line  A  B  as  radius,  describe  arcs 
cutting  the  perpendiculars  A  E  and  B  F  in  E  and  F,  and  join  G  E, 
E  F,  F  D,  to  complete  the  octagon. 

Otherwise,  thus. — Let  A  B  (Fig.  36)  be  the  given  straight  line  on 
which  the  octagon  is  to  be  described.  Bisect  it  in  a,  and  draw 
the  perpendicular  a  b  equal  to  A  a  or  B  a.  Join  A  l>,  and  produce 
a  b  to  c,  making  b  c  equal  to  A  b  ;  join  also  A  c  and  B  c,  extending 
them  so  as  to  make  c  E  and  c  F  each  equal  to  A  c  or  B  c. 
Through  c  draw  c  c  G  at  right  angles  to  A  E.  Again,  through  the 
same  point  f,  draw  D  H  at  right  angles  to  B  F,  making  each  of  the 
lines  c  c,  c  D,  c  o,  and  c  H  equal  to  A  c  or  c  B,  and  consequently 
equal  to  one  another.  Lastly,  join  B  c,  c  D,  D  E,  E  F,  F  G,  G  H,  H  A  ; 
ABCDEFGH  will  be  a  regular  octagon  described  upon  A  B,  as 
required. 

PROBLEM  XVII. — In  a  given  square  to  inscribe  a  given  octagon. 

Let  A  B  c  D  (Fig.  37)  be  the  given  square.  Draw  the  diagonals 
A  c  and  B  D,  intersecting  each  other  in  e ;  then  from  the  angular 
points  ABC  and  D  as  centres,  with  a  radius  equal  to  half  the 
diagonal,  viz.,  A  e  or  c  e,  describe  arcs  cutting  the  sides  of  the 


27 

square  in  the  points/  g,  h,  k,  /,  m,  ;/,  o,  and  the  straight  lines  of, 
g  h,  k  /,  and  m  n,  joining  these  points  will  complete  the  octagon, 
and  be  inscribed  in  the  square  A  u  c  D,  as  required. 

PROBLEM  XVIII. — To  find  the  area  of  any  regular  polygon. 

Let  the  given  figure  be  a  hexagon;  it  is  required  to  find  its  area. 
Bisect  any  two  adjacent  angles,  as  those  at  A  and  B  (Fig.  38),  by 
the  straight  lines  A  c  and  B  c,  intersecting  in  c,  which  will  be  the 


Fig.  37. 


Fig.  38. 


Fig.  39. 


centre  of  the  polygon.  Mark  the  altitude  of  this  elementary  tri- 
angle by  a  dotted  line  drawn  from  c  perpendicular  to  the  base 
A  B;  then  multiply  together  the  base  and  altitude  thus  found,  and 
this  product  by  the  number  of  sides  :  half  gives  the  area  of  the 
whole  figure. 

Of  otherwise,  thus.  —  Draw  the  straight  line  D  E,  equal  to  six 
times,  /'.  e.,  as  many  times  A  B,  the  base  of  the  elementary  triangle, 
as  there  are  sides  in  the  given  polygon.  Upon  D  E  describe  an 
isosceles  triangle,  having  the  same  altitude  as  A  B  c,  the  elementary 
triangle  of  the  given  polygon  ;  the  triangle  thus  constructed  is 
equal  in  area  to  the  given  hexagon;  consequently,  by  multiplying 
the  base  and  altitude  of  this  triangle  together,  half  the  product  will 
be  the  area  required.  The  rule  may  be  expressed  in  other  words, 
as  follows  :  —  The  area  of  a  regular  polygon  is  equal  to  its  perimeter, 
multiplied  by  half  the  radius  of  its  inscribed  circle,  to  which  the 
sides  of  the  polygon  are  tangents. 

PROBLEM  XIX.  —  To  describe  the  circumference  of  a  circle  tlirough 
three  given  points. 

Let  A,  B,  and  c  (Fig.  39)  be  the  given  points  not  in  a  straight 
line.  Join  A  Band  B  c;  bisect  each  of  the  straight  lines  A  B  and 
B  c  by  perpendiculars  meeting  in  D  ;  then  A,  B  and  c  are  all  e?,ui- 


HAND-RAILING. 


distant  from  n ;  therefore  a  circle  described  from  T),  with  the  radius 
D  A,  DB,  or  DC,  will  pass  through  all  the  three  points  as  required. 

PROBLEM  XX. — To  divide  a  given  circle  into  any  number  of  equal 
or  proportional  parts  by  concentric  divisions. 

Let  ABC  (Fig.  40)  be  the  given  circle,  to  be  divided  into  five 
equal  parts.  Draw  the  radius  A  D,  and  divide  it  into  the  same 


Fig.  40.  Fig.  41.  Pig.  42. 

number  of  parts  as  those  required  in  the  circle;  and  upon  the 
radius  thus  divided,  describe  a  semicircle :  then  from  each  point  of 
division  on  A  D,  erect  perpendiculars  to  meet  the  semi-circumfer- 
ence in  e,f,g,  and  h.  From  D,  the  centre  of  the  given  circle, 
with  radii  extending  to  each  of  the  different  points  of  intersection 
on  the  semicircle,  describe  successive  circles,  and  they  will  divide 
the  given  circle  into  five  parts  of  equal  area  as  required ;  the  centre 
part  being  also  a  circle,  while  the  other  four  will  be  in  the  form  of 
rings. 

PROBLEM  XXI. — To  divide  a  circle  into  three  concentric  parts, 
bearing  to  each  other  the  proportion  of  one,  two,  three,  from  t/ic  centre. 

Draw  the  radius  A  D  (Fig.  41),  and  divide  it  into  six  equal  parts. 
Upon  the  radius  thus  divided,  describe  a  semicircle :  from  the  first 
and  third  points  of  division,  draw  perpendiculars  to  meet  the  semi- 
circumference  in  e  and/  From  D,  the  centre  of  the  given  circle, 
with  radii  extending  to  e  and/,  describe  circles  which  will  divide 
the  given  circle  into  three  parts,  bearing  to  each  other  the  same 
proportion  as  the  divisions  on  A  D,  which  are  as  i,  2  and  3.  In 
like  manner  circles  maybe  divided  in  any  given  ratio  by  concentric 
divisions. 


RAND-RAILING.  29 

PROBLEM  XXII. — To  draw  a  straight  line  equal  to  any  given  are 
of  a  circle. 

Let  A  B  (Fig.  42)  be  the  given  arc.  Find  c  the  centre  of  the  arc, 
and  complete  the  circle  A  D  B.  Draw  the  diameter  B  D,  and  pro- 
duce it  to  E,  until  D  E  be  equal  to  c  D.  Join  A  E,  and  extend  it  so 
as  to  meet  a  tangent  drawn  from  B  hi  the  point  F  ;  then  B  F  will 
be  nearly  equal  to  the  arc  A  B. 

The  following  method  of  finding  the  length  of  an  arc  is  equally 
simple  and  practical,  and  not  less  accurate  than  the  one  just 
given. 

Let  A  B  (Fig.  43)  be  the  given  arc.  Find  the  centre  c,  and  join 
A  B,  B  c,  and  CA.  Bisect  the  arc  A  B  in  D,  and  join  also  c  D; 
then  through  the  point  D  draw  the  straight  line  E  D  F,  at  right 


Fig.  43.  Fig.  44. 

angles  to  c  D,  and  meeting  c  A  and  c  B  produced  in  E  and  F.  Again, 
bisect  the  lines  A  E  and  B  F  in  the  points  G  and  H.  A  straight  line 
G  H,  joining  these  points,  will  be  a  very  near  approach  to  the  length 
of  the  arc  A  B. 

Seeing  that  in  very  small  arcs  the  ratio  of  the  chord  to  the  double 
tangent  or,  which  is  the  same  thing,  that  of  a  side  of  the  inscribed 
to  a  side  of  the  circumscribing  polygon,  approaches  to  a  ratio  of 
equality,  an  arc  may  be  taken  so  small,  that  its  length  shall  differ 
from  either  of  these  sides  by  less  than  any  assignable  quantity ; 
therefore,  the  arithmetical  mean  between  the  two  must  differ  from 
the  length  of  the  arc  itself  by  a  quantity  less  than  any  that  can  be 
assigned.  Consequently  the  smaller  the  given  arc,  the  more  nearly 
will  the  line  found  by  the  last  method  approximate  to  the  exact 
length  of  the  arc.  If  the  given  arc  is  above  60  degrees,  or  two- 
thirds  of  a  quadrant,  it  ought  to  be  bisected,  and  the  length  of  the 


30  HAND-RAILINC,. 

semi-arc  thus  found  being  double,  will  give  the  length  of  the  whole 
arc. 

These  problems  arc  very  useful  in  obtaining  the  lengths  of  veneers 
or  other  materials  required  for  bending  round  soffits  of  door  and 
window-heads. 

PROBLEM  XXIII. —  To  describe  the  segment  of  a  circle  by  means 
of  two  laths,  the  chord  and  versed  sine  being  given. 

Take  two  rods,  E  B,  B  F  (Fig.  44),  each  of  which  must  be  at  least 
equal  in  length  to  the  chord  of  the  proposed  segment  AC:  join 
them  together  at  B,  and  expand  them,  so  that  their  edges  shall  pass 
through  the  extremities  of  the  chord,  and  the  angle  where  they  join 
shall  be  on  the  extremity  B  of  the  versed  sine  D  B,  or  height  of  the 
segment.  Fix  the  rods  in  that  position  by  the  cross  piece  g  /i, 
then  by  guiding  the  edges  against  pins  in  the  extremities  of  the 
chord  line  A  c,  the  curve  ABC  will  be  described  by  the  point  B. 

PROBLEM  XXIV. — Having  the  chord  and  versed' sine  of  the  seg- 
ment of  a  circle  of  large  radius  given,  to  find  any  number  of  points  i;i 
tJie  curve  by  means  of  intersecting  lines. 

Let  AC  be  the  chord  and  D  B  the  versed  sine. 

Through  B  (Fig.  45)  draw  E  F  indefinitely  and  parallel  to  A  c; 
join  A  B,  and  draw  A  E  at  right  angles  to  A  B.  Draw  also  A  o  at 


right  angles  to  A  c,  or  divide  A  D  and  E  B  into  the  same  number  of 
equal  parts,  and  number  the  divisions  from  A  and  E  respectively, 
and  join  the  corresponding  numbers  by  the  lines  i  i,  2  2,  3  3. 
Divide  also  A  G  into  the  same  number  of  equal  parts  as  A  L>  or  K  B, 
numbering  the  divisions  from  A  upwards,  1,2,  3,  etc.;  and  from  the 
points  ,  2  and  3,  draw  lines  to  B  ;  and  the  points  of  intersection  of 
these,  with  the  other  lines  at  //,  k,  /,  will  be  points  in  the  curve  re- 
quired. Same  with  r,  c. 

Anotlier  Method. — Let  AC  (Fig.  46)  be   the  chord   and  D  B  the 
versed  sine.     Join  A  B,  B  c,  and  through  i;  dra\v  E  F  parallel  to  AC. 


HAND- RAILING.  3! 

From  the  centre  B,  with  the  radius  B  A  or  B  c,  describe  the  arcs 
A  E,  c  F.  and  divide  them  into  any  number  of  equal  parts,  as  i,  2, 
3:  from  the  divisions  i,  2,  3,  draw  radii  to  the  centre  B,  and  divide 
each  radius  into  the  same  number  of  equal  parts  as  the  arcs  A  E 


Fig.  46. 

and  c-  F  ;  and  the  points  g,  //,  /,  ///,  //,  o,  thus  obtained,  are  points  in 
the  required  curve. 

These  methods,  though  not  absolutely  correct,  are   sufficiently 
accurate  when  the  segment  is  less  than  the  quadrant  of  a  circle. 

PROBLEM  XXV. — To  draw  an  ellipse  with  the  trammel. 

The  trammel  is  an  instrument  consisting  of  two  principal  p.-irts, 
the  fixed  part  in  the  form  of  a  cross  E  F  GH  (Fig.  47),  and  the 
moveable  piece  or  tracer  kl m.  The  fixed  piece  is  made  of  two 
rectangular  bars  or  pieces  of  wood,  of  equal  thickness,  joined  to- 
gether so  as  to  be  in  the  same  plane.  On  one  side  of  the  frame 
so  formed,  a  groove  is  made,  forming  a  right-angled  cross.  In  the 
groove  two  studs,  k  and  /,  are  fitted 
to  slide  freely,  and  carry  attached  to 
them  the  tracer  k  I  m.  The  tracer 
is  generally  made  to  slide  through  a 
socket  fixed  to  each  stud,  and  pro- 
vided with  a  screw  or  wedge,  by 
which  the  distance  apart  of  the  studs 
may  be  regulated.  The  tracer  has 
another  slider  ;//,  also  adjustable, 
which  carries  a  pencil  or  point.  The  instrument  is  used  as  fol- 
lows : — Let  A  c  be  the  major,  and  H  B  the  minor  axis  of  an  ellipse : 
lay  the  cross  of  the  trammel  on  these  lines,  so  that  the  centre  lines  of 
it  may  coincide  with  them ;  then  adjust  the  sliders  of  the  tracer,  so 
that  the  distance  between  k  and  ///  may  be  equal  to  half  the  major 
axis,  and  the  distance  between  /  and  >n  equal  to  half  the  m'inr 


Fig.  47. 


32 


HAND-RAILING. 


Fig.  48. 


axis;  then  by  moving  the  oar  round,  the  pencil  in  the  slider  will 
describe  the  ellipse. 

PROBLEM  XXVI. — An  ellipse  may  also  be  described  by  means  of  a 
string. 

Let  A  B  ( Fig.  48)  be  the  major  axis,  and  D  c  the  minor  axis  of 
the  ellipse,  and  F  G  its  two  foci. 
Take  a  string  EOF  and  pass  it 
over  the  pins,  and  tie  the  ends, 
together,  so  that  when  doubled  it 
may  be  equal  to  the  distance  from 
the  focus  F  to  the  end  of  the  axis, 
B;  then  putting  a  pencil  in  the 
bight  or  doubling  of  the  string  at 
H  and  carrying  it  round,  the  curve 
may  be  traced.  This  is  based  on 
the  well  known  property  of  the 
ellipse,  that  the  sum  of  any  two 
lines  drawn  from  the  foci  to  any 
points  in  the  circumference  is  the  same. 

PROBLEM  XXVII. — The  axes  of  an  ellipse  being  given,  to  draw 
the  curve  by  intersections. 

Let  A  c  (Fig.  49)  be  the  major  axis,  and  D  B  half  the  minor 
axis.  On  the  major  axis  construct  the  parallelogram  AEFC,  and 
make  its  height  equal  to  D  B.  Divide  A  E  and  E  B  each  into 
the  same  number  of  equal  parts,  and  number  the  divisions  from  A 
and  E  respectively;  then  join  A  i,  i  2,  2  3,  etc.,  and  their  intersec- 
tions will  give  points  through  which  the  curve  may  be  drawn. 

Th«  points  for  a  "  raking  "  or  rampant  ellipse  may  also  be  found 
by  the  intersection  of  lines  as  shown  at  Fig.  50.  Let  A  c  be  the 
major  and  E  B  the  minor  axis :  draw  A  G  and  c  H  each  parallel 
to  B  E,  and  equal  to  the  semi-axis  minor.  Divide  A  i),  the  semi- 
axis  major,  and  the  lines  A  G  and  c  H  each  into  the  same  number 
of  equal  parts,  in  i,  2,  3  and  4;  then  from  E,  through  the  divisions 
i,  2,  3  and  4,  on  the  semi-axis  major  A  D,  draw  the  lines  E  //,  E  k,  E  /, 
and  EOT;  and  from  n,  through  the  divisions  i,  2,  3  and  4  on  the 
line  A  G,  draw  the  lines  i,  2,  3  and  4  is;  and  the  intersection  of 


HAND- KAN.  I  Mi. 


Fig..  49.  Fig.  50. 

these  with  the  lines  E  i,  2,  3  and  4  in  the  points  h  klm,  will  be 
points  in  the  curve. 

PROBLEM  XXVIII. — 7o  describe  with  a  compass  a  figure  resem- 
bling the  ellipse. 

Let  A  B  (Fig.  51)  be  the  given  axis,  which  divide  into  three  equal 
parts  at  the  points/^.  From  these  points  as  centres,  with  the  radius 
/A,  describe  circles  which  intersect  each  other,  and  from  the  points 
of  intersection  through  /and  g,  draw  the  diameters  CgE,  c/n. 
From  c  as  a  centre,  with  the  radius  c  D,  describe  the  arc  D  F,  which 


Fig.  51.  Fig.  52. 

completes  the  semi-ellipse.  The  other  half  of  the  ellipse  may  be 
completed  in  the  same  manner,  as  shown  by  the  dotted  lines. 

PROBLEM  XXIX. — Another  method  of  describing  a  figure  ap- 
proaching the  ellipse  with  a  compass. 

The  proportions  of  the  ellipse  may  be  varied  by  altering  the 
ratio  of  the  divisions  of  the  diameter,  as  thus  : — Divide  the  major 
axis  of  the  ellipse  A  B  (Fig.  52),  into  four  equal  parts,  in  the  points 
f  g  h.  On/// construct  an  equilateral  triangle /c //,  and  produce 


34 


HAND-RAILING. 


HAND-RAILING.  35 

- ;  the  sides  of  the  triangle  c  /,  c  //  indefinitely,  as  to  D  and  E.  Then 
froiii  the  centres  f  and  h,  with  the  radius  A  f,  describe  the  circles 
A  D  g,  B  E  g;  and  from  the  centre  c,  with  the  radius  c  D,  describe 
the  arc  D  E  to  complete  the  semi-ellipse.  The  other  half  may  be 
completed  in  the  same  manner.  By  this  method  of  construction 
the  minor  axis  is  to  the  major  axis  as  14  to  22. 

The  following  problems,  which  relate  more  particularly  to  hand- 
railing,  should  be  thoroughly  mastered  before  passing  to  actual 
work. 

A  tangent  is  a  line  touching  a  circle  at  right  angles  to  the  radius 
as  shown  at  Fig.  55. 

To  construct  Fig.  53 :  From  the  centre  o  with  the  radius  o  A, 
describe  a  quarter-circle  A  p  c;  draw  tangents  A  B  and  c  B;  join 
A  c;  through  the  point  B  draw  a  straight  line  parallel  to  A  c;  with 
the  centre  B,  with  the  radius  B  A,  describe  the  arcs  A  D  and  c  E;  at 
the  point  E  erect  the  perpendicular  E  F  at  right  angles  to  n  E  to 
any  desired  height;  in  laying  down  a  hand-rail  this  height  will  be  the 
number  of  risers  contained  in  the  wreath;  let  F  be  the  given  height 
(this  being  one  pitch);  join  F  D;  extend  o  B  to  G;  draw  GH  at  right 
angles  to  F  D  ;  make  G  H  equal  to  B  i ;  with  the  centre  H  and  ra- 
dius D  G  describe  arcs  cutting  D  F  at  K  and  L;  draw  H  L  and  H 
K,  which  are  the  tangents  on  the  pitch,  and  which  when  placed  in 
position  would  stand  plumb  over  ABC. 

To  construct  Fig.  54,  proceed  as  above  until  the  height  is 
located.  It  will  be  seen  that  here  B  G  is  lifted  higher,  making  the 
pitch  line  and  tangents  F  G  and  D  G  of  unequal  length.  To  obtain 
the  angle,  continue  B  G  to  H,  making  B  H  equal  to  E  F  ;  from  H 
draw  the  line  H  j  to  any  distance  at  right  angles  to  D  G.  With  the 
centre  G  and  radius  G  F,  describe  an  arc  cutting  the  line  H  J  at  i ; 
join  i  G  and  i  D,  and  the  angle  is  completed. 

An  easy  way  to  prove  these  problems  is  to  draw  them  on  com- 
mon thick  paper ;  then  take  a  knife,  and  cut  out  the  angle  D  E  F, 
place  it  perpendicularly  over  ABC,  bringing  D  over  A,  and  E  over 
c;  then  cut  out  the  angle  H  L  K,  and  if  drawn  correctly  it  will  lie  on 
the  pitch  lines  and  fit  the  sides  exactly. 

To  draw  the  curve  line  the  quickest  and  most  practical  method 


HAND-RAILING. 


Fig.  54. 


HAND-RAILING. 


37 


is  to  take  B  as  a  centre,  and  with  a  radius  of  B  p,  describe 
an  arc,  touching  the  curve  A  p  c,  in  the  angle  ABC,  from  H,  as  a 
centre,  with  the  same  radius,  describe  an  arc  cutting  H  G  ai  M  ; 
then,  take  a  thin  flexible  strip  of  wood  of  an  even  thickness,  bend  it 
until  it  touches  the  points  KIM;  mark  around  it  with  a  pencil, 
and  the  curve  is  completed,  and  near  enough  to  absolute  accuracy 
tor  ;ill  practical  purposes. 


Fig.  55. 


38  HAND-RAIUNG. 


PART  II. 

TO  MAKE  THE  WREATHS  FOR  STAIRS  WITH   FOUR 
WINDERS  IN  THE  WELL. 

FIGURE  i. — Strike  the  quarter  circle  ABC,  which  is  the  centre 
line  of  the  rail,  and  also  describe  the  inside  and  outside  lines  of  rail 
b  b  l>,  and  mark  the  risers  at  A  B  and  C.  Draw  the  lines  A  D 
and  CD. 

FIGURE  2. — Draw  the  straight  line  B  K  equal  to  the  tread  of  one 
step,  and  set  up  K  A  equal  to  the  rise  of  one  step.  Draw  the  line 
A  C,  and  make  A  D  and  D  C  equal  to  the  length  A  D  and  D  C 
in  Figure  i.  Set  up  the  height  of  two  risers  from  C  to  F.  Draw 
the  line  A  B,  and  mark  the  point  G  at  one  third  the  distance  from 
A  to  B,  and  draw  the  line  F  G.  Continue  C  A  to  the  point  H, 
and  from  D  draw  the  perpendicular  line  D  J.  Make  the  point  at 
H square  to  G  H,  and  ease  off  the  rail  from  /to  H,  and  draw  the 
top  and  bottom  side,  by  setting  half  the  thickness  of  the  rail  each  side 
the  line  /  H. 

FIGURE  i. — Draw  the  line  C I  through  the  centre  X,  and  make 
C I  equal  to  D  H  in  Figure  2,  and  draw  HI.  Set  up  C  F  equal 
to  C  F  in  Figure  2,  and  C  J  equal  to  D  J  in  Figure  2.  Join 
J  I,  and  from  the  point  fdraw  the  line  f  G  parallel  to  J  I,  and 
join  G  H.  Draw  ordinate  lines  a  b  b  b  at  any  convenient  dis- 
tance apart  all  parallel  to  G  H,  one  of  which  to  pass  from  D  to  X. 
From  the  point  /draw  the  line  IN,  square  to  J  /,  and  from  G 
with  the  distance  G  H  cut  the  line  /,  N  at  E,  and  draw  G  E. 
Draw  the  perpendicular  lines  a  c  and  from  the  points  c  c  c  draw 
ordinate  lines  c  d,  all  parallel  to  G  E.  Take  the  distance  from  a 
to  b  b  b,  and  apply  them  from  c  to  dt  d,  dt  and  draw  the  face  mould 


HAND-RAILING. 


39 


Fig.  L 


HAND-RAILING. 


Fig.  2. 


HAND-RAILING. 


42  HAND-RAILING. 

through  them.  Take  the  distance  X  D,  and  mark  it  on  the  ordi- 
nate  from  c  to  k,  and  draw  the  lines  E  K  and  K  F.  Make  the 
joint  L  M  square  to  K  F,  and  the  end  N  O  square  to  K  E. 
From  /,  with  the  distance  I  P,  describe  the  arc  P  Q,  and  draw 
Q  If,  and  apply  this  bevel  I  Q  Jf,  to  both  ends  of  the  wreath 
through  the  centre  O,  as  shown  at  /  Q  H  in  Figure  4.  Take  the 
distance  Q  O  in  Figure  4,  and  apply  it  from  /  to  ft  in  Figure  i ; 
draw  J?  S  parallel  to  /  J,  and  cut  it  with  the  line  /,  S.  Apply 
the  distance  R  S  from  F\.o  V,  and  F  to  IV;  also  from  E  to  7' and 
E  to  U,  and  make  the  face  mould  the  length  from  7"  to  W. 

FIGURE  3. — The  centre  line  E  U  F,  and  the  joints  are  the  same 
as  E  U F'\\\  Figure  i.  Set  the  compasses  to  half  the  thickness  of 
of  the  plank  C  D  in  Figure  3^,  and  mark  it  each  side  the  centre 
line  E  U  F,  and  make  another  face  mould  to  this  (Figure  3),  mark 
it  on  the  plank,  and  cut  it  out  square,  and  joint  the  ends  square. 
Now  apply  the  face  mould  Figure  i  on  the  top  side  of  wreath, 
with  the  points  V  and  T  over  the  points  Q  Q  at  the  ends  of 
wreath,  and  mark  both  edges;  apply  it  on  the  underside  with  the 
points  W  and  U  over  the  points  H  H,  at  the  ends  of  wreath,  and 
mark  both  edges.  Work  the  inside  of  the  wreath  with  a  round  plane 
to  fit  the  rail  on  the  plan,  then  work  off  the  outside  to  the  marks 
on  the  top  and  bottom,  and  the  bevels  at  the  ends.  To  work 
the  top :  At  the  mark  K  P  on  the  face  mould,  take  an  equal  por- 
tion off  the  top  and  bottom  side  of  the  wreath  that  will  leave  it  the 
thickness  of  the  rail,  and  work  it  to  the  bevels  marked  on  the  ends, 
gradually  twisting  from  end  to  end.  The  bottom  part  can  be 
gauged  from  the  top  side. 


HAND-RAILING.  43 


TO    MAKE   THE    WREATHS    FOR    STAIRS  WITH    SIX 
WINDERS  IN  THE  WELL. 

FIGURE  4. — Strike  the  quarter  circle  A  C,  which  is  the  centre 
line  of  the  rail,  and  also  describe  the  inside  and  outside  lines  of  rail, 
bob.  Draw  the  lines  A  D,  and  D  C. 

FIGURE  5. — Draw  the  straight  line  B  K  equal  to  the  tread  of 
one  step,  and  set  up  K  A  equal  to  the  rise  of  one  step.  Draw  the 
line  A  C,  and  make  A  D  and  D  C  equal  to  A  D  and  D  C  in 
Figure  4;  setup  the  height  of  3  risers  from  C  to  F  \  draw  the  line 
A  B,  and  mark  the  point  G  at  one-third  the  distance  from  A  to  B, 
and  draw  the  line  G  F;  erect  the  perpendicular  line  D  J.  Joint 
the  rail  square  at  If,  (which  eases  the  rail  better  than  if  jointed  at 
the  spring  of  the  well  at  A).  Ease  off  the  rail  from  H  to  E,  and 
draw  If  I  parallel  to  A  C. 

FIGURE  4. — Draw  the  line  C  G  through  the  centre  X,  and 
make  C  I  equal  to  L  H'\\\  Figure  2,  and  draw  I  H.  Set  up  C  F 
equal  to  I  F  in  Figure  5,  and  C  J  equal  to  L  J;  join  J  /,  and 
from  the  point  ^draw  the  line  F  G  parallel  to  J  I,  and  join  G  H. 

Draw  ordinate  lines  a  b  b  b,  at  any  convenient  distance  apart, 
all  parallel  to  G  H,  one  of  which  to  pass  from  D  to  X.  From  the 
point  /draw  the  line  /^square  to  /  J,  and  from  G,  with  the  dis- 
tance G  H,  cut  the  line  I  N z\.  E,  and  draw  G  E. 

Draw  the  perpendicular  lines  a  c,  and  from  the  points  c  c  c,  draw 
ordinate  line  c  to  d  d  d,  all  parallel  to  G  E.  Take  the  distances 
from  a  to  b  b  ft,  and  apply  them  from  c  to  d  d  d,  and  draw  the  face 
mould  through  them. 

Take  the  distance  X  D,  and  mark  it  on  the  ordinate  from  c 
to  X,  and  draw  the  line  E  K  and  K  F.  Make  the  joint  L  M 
square  to  K  F,  and  the  end  NO  square  to  K  E. 

From  /,  with  the  distance  /  P,  describe  the  arc  PQ,  and  draw  Q 
If.  Apply  the  bevel  /  Q  H  to  both  ends  the  wreath  through  the 


44 


HAND-RAILING. 


Kg.  4. 


HAND- RAILING. 


45 


Fitf.  5. 


HAND-P  'HUNG. 


HAND-RAILING.  47 

centre  O,  as  shown  at  /  Q  H  in  Figure  4.  Take  the  distance  Q 
O  in  Figure  4,  and  apply  it  from  /  7?  in  Figure  i,  and  draw  R  S 
parallel  to  G  F,  and  cut  it  with  the  line  /  S.  Apply  the  distance 
R  .9 from  Flo  V,  and  F  io  W,  also  from  E  to  rand  E  to  U,  and 
make  the  face  mould  from  7"  to  W. 

FIGURE  6. — The  centre  line  E  U  F  and  the  joints  at  the  ends, 
are  the  same  as  E  U  F  in  Figure  i.  Set  the  compasses  to  half  the 
thickness  of  the  plank  CD  in  Figure  7,  and  mark  it  each  side  the 
centre  line  E  U  F.  Make  another  face  mould,  (to  Figure  6)  mark 
it  on  the  plank,  cut  it  out  square,  and  joint  the  ends  square. 

Apply  the  face  mould,  Figure  4,  on  the  top  side  the  wreath,  with 
the  points  T^and  Fover  the  points  Q  Q  at  the  ends  of  the  wreath, 
and  mark  both  edges.  Apply  it  on  the  underside  the  wreath,  with 
the  points  W  and  U  over  the  points  //  H  at  the  ends  of  the 
wreath,  and  mark  it. 

Work  the  inside  the  wreath  with  a  round  plane  to  fit  the  inside 
curve  on  plan,  then  work  off  the  outside  to  the  marks  on  the  top 
and  bottom  and  the  bevels  at  the  ends. 

To  work  the  top,  at  the  mark  K  P  on  the  face  mould,  take  an 
equal  portion  off  the  top  and  bottom  side  the  wreath,  that  will 
leave  it  the  thickness  of  the  rail,  and  work  it  from  the  bevels  at 
each  end,  through  K  P,  gradually  twisting  from  end  to  end.  The 
bottom  part  can  then  be  gauged  from  the  top  side. 


48  HAND-RAILING. 


TO  MAKE  THE  WREATHS  FOR  STAIRS  WITH  THREE 
WINDERS  IN  HALF  THE  WELL,  AND  LANDING 
IN  THE  CENTRE  OF  THE  WELL. 

(THIS  is  ONE  OF  THE  MOST  DIFFICULT  RAILS  THAT  CAN  BE  MADE.) 

FIGURE  8. — Strike  out  the  semi-circle  A  B  C,  which  is  the  centre 
of  rail,  and  also  describe  the  inside  and  outside  lines  of  rail.  Draw 
the  lines  C  £,  A  £>,  and  D  E. 

FIGURE  9  {See  Folding  Plate). — Draw  the  straight  line  OK  equal 
to  the  tread  of  one  step,  and  set  up  KA  equal  to  the  rise  of  one  step. 
Draw  the  line  A  E,  and  make  A  D — D  B  and  B  E  equal  to  A  D — 
D  B  and  BE  in  Figure  8.  Set  up  the  height  of  3  risers  from  E  to 
M,  and  six  inches  more  from  M  to  C,  (this  makes  the  rail  3  feet 
high  on  the  landing,  or  six  inches  higher  than  the  rail  is  on  the 
stairs,  which  should  be  2^  feet  plumb  with  the  face  of  the  risers.) 

Draw  the  line  A  O,  and  mark  the  point  G  about  one-third  the 
distance  from  A  to  O,  and  join  C  G.  Make  the  joint  at  Zf  which 
is  a  little  above  the  spring  of  the  well  at  A,  and  ease  off  the  rail 
as  shown  at  G. 

Draw  the  line  H L  /parallel  to  A  B,  and  erect  the  perpendicu- 
lar lines  L  J  and  I  F.  Now  from  the  base  line  H  L  I  and  the 
height  L  <7and  If  get  the  face  mould  and  work  the  bottom  wreath, 
as  shown  and  described  in  Figure  5,  from  the  base  l:ne  H  L  I 
and  the  heights  L  J"and  IF. 

To  work  the  top  wreath  proceed  as  follows,  viz. : — 

FIGURE  10. — Lay  down  the  plan  of  the  rail,  the  centre  line  B 
C  F,  being  the  same  as  B  C  F,  in  Figure  8,  and  make  the  straight 
part  about  5  inches  from  C  to  F.  Draw  the  lines  F  A-A  B-B  E 
and  E  F.  Take  the  height  N  C  in  Figure  9,  and  apply  it  from  F 
to  D,  and  join  A  D.  Draw  D  G  square  to  A  D,  and  make  D  G 
equal  to  F  E. 


HAND-RAILING. 


49 


fig.  10. 


.  8. 


5° 


HAND-RAILING. 


M 


HAND-RAILING.  51 

Draw  ordinate  lines  a  b  b  b  parallel  to  A  B,  and  draw  the  lines 
a  c,  and  from  c  c,  draw  ordinate  c  d  d  d,  parallel  to  G  D.  Apply 
the  distance  from  a  to  b,  b,  b,  on  to  the  ordinates  from  c  to  d,  d,  d, 
and  draw  the  face  mould  through  them.  Make  the  joints  at  H 
and  D  square,  as  shown. 

FIGURE  n. — Shows  the  ends  of  the  wreath.  Take  the  bevel 
F  D  A  in  Figure  10  and  apply  it  at  the  top  end  of  the  wreath 
through  the  centre  O  at  F  D  A,  and  mark  the  size  of  the  rail  on 
as  shown. 

FIGURE  12. — Is  the  face  mould  to  cut  the  wreath  out  square;  the 
centre  line  DIJf,  and  the  joints  are  the  saifie  as  the  centre  line  D 
I H'm  Figure  3;  put  half  the  thickness  of  the,plank  each  side  the 
centre  line  D  Iff,  and  cat  out  the  wreath.  Figure  8:  Take  half  the 
thickness  of  the  plank  from  A  to  Z,  draw  L  M  parallel  to  A  D,  and 
draw  A  M,  apply  the  distance  L  J/to  //  J,  and  H  K,  and  make 
the  face  mould  from  J  to  D. 

Apply  Figure  10  face  mould  on  the  top  side  of  wreath,  wiih  the 
points  D  and  J,  over  the  points  D  J  in  Figure  n,  and  mark  it. 
Apply  it  on  the  underside  with  the  points  D  and  .AT  over  the  points 
A  and  K  in  Figure  n. 

Work  it  as  described  in  the  preceding  Figures  6  and  7,.  to  th« 
bevels  at  the  ends  as  shown,  and  work  the  part  from  D  to  /straight; 
then  work  the  top  gradually  twisting  from  /to  H  in  Figure  12. 


.    t 


HAN  iv  RAILING. 


TO    MAKE   THE  WREATHS    FOR  STAIRS  WITH    ONE 
RISER  IN  THE  CENTRE  OF  WELL. 

FIGURE  13  (See Folding  Plate}. — Mark  the  plan  of  the  rail  as  A  B 
CD,  and  E F  G  the  centre  line,  the  straight  part  E F  being  the  width 
of  one  step.  Continue  E  F  to  H  and  draw  G  H.  Draw  D  /paral- 
lel to  E  F,  and  draw  the  lines  A  J  and  F  0.  Make  0  K  equal  to  the 
rise  of  one  step,  and  draw  J  K,  set  up  G  L  equal  to  the  rise  of  two 
steps.  From  the  point  L  make  the  line  L  /parallel  to  J  K  and 
join  /  E.  Draw  ordinate  lines  as  i,  2,  3,  4,  at  any  convenient 
distance  apart  all  parallel  to  /  E,  draw  the  perpendicular  lines  i, 
5,  etc.,  to  intersect  the  line  I  L.  From  the  point  J,  draw  the  line 
J M,  square  to  J  K,  and  from  7,  with  the  distance  IE,  cut  J M at 
the  point  N,  and  draw  /  N;  draw  ordinate  lines  5,  6,  7,  8,  etc.,  all 
parallel  to  /  N.  Take  the  distances  i,  2,  3,  4,  and  apply  them 
from  5  to  6,  7,  8,  etc.,  and  draw  the  inside  and  outside  and  centre 
line  of  face  mould  through  them. 

Take  the  length  of  the  ordinate  0  H,  and  apply  it  on  the  ordinate 
from  P  to  Q,  and  draw  the  lines  N  Q,  and  L  Q.  Make  the  joint  R 
8  square  to  Q  L,  and  the  joint  M  N  square  to  Q  N. 

FIGURE  14. — The  centre  line  N  7  7  7  L,  and  the  joints  at  ends, 
are  the  same  as  Figure  13. 

Set  the  compasses  to  half  the  thickness  of  the  plank,  (as  shown 
at  G  D  in  Figure  15),  and  apply  it  on  each  side  the  centre  line 
L  7  7  7  JV,  and  draw  the  inside  and  outside  lines.  Mark  this  face 
mould  on  the  plank,  and  cut  it  out  square,  and  joint  Lhe  ends 
square. 

To  put  the  twist  on  the  wreath,  proceed  as  follows : — 

In  Figure  13  from  the  centre  J,  describe  the  arc  T  U,  and  draw 
U  E.  Continue  the  lines  /  E,  and  G  H,  until  they  meet  at  W. 
Take  the  length  G  L  and  apply  it  from  G  to  V,  and  draw  V  W. 
From  G  draw  G  X  square  to  V  W,  and  describe  the  arc  X  Y,  and 
draw  YI. 


HAND-RAILING. 


53 


54  HAND-RAILTNG. 

Take  the  bevel  U  J  E,  and  apply  it  on  the  bottom  end  of  the 
wreath  in  Figure  15,  at  U  J  E,  through  the  centre  O,  and  mark 
half  the  width  of  the  rail  on  each  side  U  E,  also  mark  the  top  and 
bottom  line  of  rail  square  to  U  E.  Apply  the  bevel  H  Y  I'm  Fig- 
ure i  to  H  Y  I,  at  the  top  end  of  the  wreath  in  Figure  15,  through 
the  centre  0,  and  mark  the  size  of  the  rail  on  as  shown.  Mark 
the  line  Q  P  on  both  sides  the  wreath.  Take  the  distance  0  U 
in  Figure  15,  and  apply  it  from  J  to  a  in  Figure  13,  draw  a  b  par- 
allel to  J  K,  and  draw  J  b,  apply  the  distance  a  b  from  N  to  c  and 
N  to  d.  Apply  the  distance  0  Y,  in  Figure  15,  to  X  e  in  Figure 
13,  and  draw  ^/parallel  to  G  V,  and  apply  the  distance  X  f  to  L 
g,  ami  L  h.  Make  a  face  mould  to  Figure  13,  the  full  length 
from  c  to  h,  and  place  it  on  the  top  of  the  wreath,  with  the  point 
^over  the  point  Y,  in  Figure  15,  at  the  top  end  of  the  wreath,  and 
the  point  c  over  the  point  U,  at  the  bottom  end  and  mark  both 
edges,  then  mark  it  on  the  underside  the  wreath,  with  the  point  h 
over  the  point  /  in  Figure  15,  and  the  point  c  over  the^oint  E. 

First  work  off  the  inside  of  the  wreath  with  a  round  plane  that 
will  fit  the  rail  on  the  plan,  then  work  off  the  outside  part.  To  work 
the  top  at  the  mark  Q  P,  take  an  equal  portion  off  the  top  and  bot- 
tom side  of  the  wreath  that  will  leave  it  the  finished  thickness  of  the 
rail,  then  work  from  the  bevels  marked  at  each  end,  through  the 
part  taken  off  at  Q  P  and  the  bottom  part  can  be  gauged  from 
the  top. 


HAND-RAILING.  55 


TO  MAKE  THE  WREATHS  FOR  LEVEL  LANDING 
STAIRS.  THE  DISTANCE  BETWEEN  CENTRE  OF 
RAILS  ACROSS  THE  WELL  BEING  EQUAL  TO 
THE  TREAD  OF  A  STEP,  OR  NEARLY  SO. 

FIGURE  16. — A  B  C  is  the  centre  of  tlie  rail,  A  B  being  equal  to 
the  tread  of  one  step.  Strike  out  the  inside  and  outside  of  the  rail  b 
b  b  b.  Draw  C  D  and  D  A.  Set  up  E  F  equal  to  one  riser,  and 
draw  D  F  to  G,  and  draw  the  line  C  G.  Draw  ordinate  lines 
a  b  b,  at  any  convenient  distance  apart,  and  draw  the  lines  tf,  c. 
From  the  points  c  c  draw  ordinate  lines  c  d.  square  to  the  line  D  G. 
Take  the  distance  a  b  b,  and  apply  them  on  the  ordinates  from  c 
to  d  d,  and  draw  the  face  mould  through  them. 

Take  the  thickness  of  the  plank  C  A  in  Figure  18,  and  apply  it 
from  D  to  /  in  Figure  16,  draw  I H  parallel  to  D  F,  and  cut  it  with 
the  line  of  D  H.  Take  half  of  /  H  and  apply  it  from  K  to  L,  and 
K  to  J.  Make  the  face  mould  the  length  from  J  to  G,  and  mark 
it  on  the  top  side'  of  the  plank,  and  cut  the  two  joints  at  J  and  G 
square. 

Take  the  bevel  D  J  E  in  Figure  16,  and  apply  it  to  the  top  joint 
at  D  F  E,  as  shown  in  Figure  18.  The  bottom  end  will  be  square, 
as  shown  in  Figure  17. 

Apply  the  face  mould  underside  the  wreath  with  the  point  L 
at  the  lower  end  of  wreath.  Cut  it  out  bevel  (as  this  kind  ol 
wreath  is  less  labor  cut  bevel  than  having  to  bevel  it  if  cut  out 
square). 

Take  an  equal  portion  off  the  top  and  bottom  side  of  the  shank- 
part  of  wreath  to  leave  the  rail  the  finished  thickness  as  shown  in 
Figure  17,  at  the  bottom  end,  then  gradually  work  the  top  side 
twisted  to  the  bevel  at  the  top  end,  so  that  when  the  wreath  is 
put  up  to  its  proper  pitch  the  top  side  is  level  across.  Gauge  the 
bottom  from  the  top  side. 


HAND-RAILING. 


Fig.  16. 


HAND-UALL1NG. 


57 


I 


HAND-RAILING. 


HAND-RAILING.  59 

NOTE. — The  distance  across  the  well  between  centre  of  rails 
should  not  exceed  the  tread  of  a  step  for  level  landing  stairs,  or  the 
wreath  has  the  appearance  of  dropping. 

If  the  distance  is  less  than  the  tread  of  a  step  bevel  ordinate 
lines  would  be  required,  after  the  manner  shown  in  Figures  13,  14, 
and  .15. 


60  HAND-RAILING. 


TO   MAKE  THE   WREATH  FOR   QUARTER-LANDING 
STAIRS,  HAVING  ONE   WINDER  IN  THE  WELL. 

FIGURE  19. — From  the  centre  0  strike  the  quarter  circle  A  Jj, 
the  centre  of  the  rail,  and  draw  the  lines  0  A — 0  B — B  C  and  A 
C,  and  draw  the  inside  and  outside  curve  of  rail. 

FIGURE  20  (See  Folding  Plate).— Make  D  E  equal  to  the  tread 
of  one  step,  and-EA  equal  to  one  riser. 

Draw  the  line  A  B  and  make  A  B  and  B  C  equal  to  A  B  and  B 
C  in  Figure  19. 

Set  up  the  height  of  2  risers  from  B  to  F,  and  draw  the  line  F 
G,  to  pitch  of  the  stairs.  Join  A  D  and  mark  the  point  H  about 
one-third  the  distance  from  A  to  Z>,  then  mark  the  point  /  about 
one-third  the  distance  from  F  to  G,  and  draw  the  line  I  H. 

Draw  the  perpendicular  line  C  J,  and  continue  A  B  to  K. 

Joint  the  end  at  K  square  to  J  K  and  ease  off  the  rail  at  H. 

Joint  the  top  length  at  the  point  L  and  ease  off  the  rail  at  /. 

FIGURE  19. — Make  B  K  equal  to  C  K  in  Figure  20.  Make  B 
J  equal  to  C  Jin  Figure  20,  and  B  M  equal  to  B  M  in  Figure  20. 
Draw  the  line  J  K,  and  from  the  point  M  draw  M  D  parallel  to 
J  K.  Draw  K  J*7  square  to  J  K,  and  from  the  point  D  with  the 
distance  D  E  cut  it  at  the  point  G.  Draw  the  ordinate  a  b  b  b 
parallel  to  D  E,  draw  the  lines  a  c,  and  from  c  draw  ordjnates  c  d  d 
d  parallel  to  D  G. 

Apply  the  distances  a  b  b  b,  on  to  the  ordinates  from  c  to  d  a  d, 
and  draw  the  face  mould  through  the  points  d  d  d. 

Mark  the  length  of  the  ordinate  0  C,  on  to  the  ordinate  from  c 
to  H,  and  draw  H  G  and  H  M.  Joint  the  end  at  G  square  to 
GH. 

Take  the  distance  M  L  in  Figure  20,  and  apply  it  from  M  to  T, 
and  joint  the  end  at  T  square  to  H  T.  Apply  the  bevel  K I  E  to 
both  ends  the  wreath  as  shown  in  Figure  22. 


HAND-RAILING. 


61 


\ 


Fiji.  19. 


62 


HAND-RAILING. 


The  face  mould  Figure  21,  is  got  from  the  centre  line  of  Figure 
19,  as  before  described,  and  the  joints  at  T  and  G  are  the  same  as  in 
Figure  19. 

Apply  the  distance  0  7,  in  Figure  22  to  K  L  in  Figure  19,  draw 
L  N,  and  mark  the  length  L  N  from  G  to  P  and  G  to  Q,  also 
from  T  to  S  and  T  to  R.  Make  the  face  mould  Figure  19,  the 
length  from  P  to  R. 

Fig.  n. 


Cut  the  wreath  out  square  to  Figure  21,  face  mould  and  joint 
the  ends.  Apply  Figure  19  face  mould  on  the  top  side  of  wreath 
with  the  points  S  P,  over  I 1  at  the  ends  of  wreath,  then  apply  it 
underside  with  the  points  R  Q,  over  E  E  at  the  ends. 

Work  the  inside  the  wreath  first  with  a  round  plan  to  fit  the  rail  on 
plan,  and  work  the  top  side  straight  from  T  to  M  and  d  to  G  in 
Figure  21,  and  gradually  work  it  twisted  to  the  bevels  at  each  end, 
and  take  an  equal  portion  off  the  top  and  bottom  side  at  c  H  to 
leave  the  rail  the  finished  thickness. 


HAND-RAILING.  63 


TO  MAKE  THE  WREATH  FOR  QUARTER  LAND- 
ING STAIRS,  WITH  THE  RISERS  AS  SHOWN  IN 
FIGURE  23. 

FIGURE  23. — From  the  centre  0  strike  the  quarter  circle  A  /?, 
which  is  the  centre  of  the  rail,  draw  the  lines  0  A — 0  B — B  C 
and  .1  C,  ai:d  draw  the  straight  part  each  end  about  3  inches  long 
from  A  to  D,  and  B  to  E,  then  draw  the  inside  and  outside  lines 
of  rail. 

FIGURE  24  (See  Folding  Plate]  — Make  D  E  equal  to  one  tread, 
and  E  A  equal  to  one  riser.  Draw  the  line  A  B,  the  parts  A  C 
and  C  B,  being  equal  to  A  C  and  C  B  in  Figure  i.  Set  up  B  F 
equal  to  one  riser  and  from  F  draw  the  line  G  F  H,  to  the  pitch  of 
the  stairs,  and  draw  (J  G. 

Join  A  D,  and  mark  the  point  /about  half  way  between  A  and 
Z>,  and  draw  /  G. 

Make  K  J"  equal  to  A  D  in  Figure  23,  and  joint  the  end  at  J, 
square  to  7  (7,  draw  the  line  J  N. 

Make  F  L  equal  to  E  B  in  Figure  23,  draw  L  M  and  joint  the 
end  at  M  square  to  F  H. 

FIGURE  23. — Draw  the  line  D  J,  set  up  B  F  equal  to  0  G  in 
Figure  24,  and  joint  J  F. 

Set  up  B  G  equal  to  N  F  in  Figure  24,  and  from  the  point  G 
draw  G  H  parallel  to  J  F. 

From  J,  draw  J I  square  to  J  F,  and  from  H  with  the  distance 
//  ft,  cut  J  /  at  K,  and  draw  If  K. 

Draw  the  ordinates  a  b  b  If,  parallel  to  H  D,  draw  the  lines  a  c, 
and  from  the  point  c,  draw  ordinate  c  d  d  d  parallel  to  H  K; 
apply  the  distance  a  b  b  b,  on  to  the  ordinates  c  d  d  d,  and  draw 
the  face  mould  through  the  points  d  d  d. 

Mark  the  length  of  the  ordinate  a  C,  on  the  ordinate  c  L,  and 
draw  L  K  and  L  <!. 


HAND- RAILING. 


Fig.  23. 


HAND-RAILING. 


Make  G  M  equal  to  F  M  in  Figure  24.  Joint  the  end  at  K 
squat -<j  to  A'  />,  and  the  end  at  M  square  to  L  M. 

Get  the  level  0  J  D  (or  the  bottom  end,  and  jB  #  0  for  the  top 
end  of  wreath,  as  described  in  Figure  15,  and  apply  them  at  the 
ends  as  shown  in  Figure  26. 

The  face  mould  Figure  25,  is.  obtained  from  the  centre  line  of 
Figure  23,  as  described  in  previous  examples. 


Fig.  26. 

Apply  the  distance  T  U  from  K  to  V,  and  K  to  W,  and  the 
length  E  Y  apply  from  M  to  g,  and  M  to  /,  and  make  Figure  23, 
face  mould  the  length  from  Fto/ 

Cut  the  wreath  out  square,  to  Figure  25,  face  mould,  and  joint 
the  ends. 

Apply  Figure  23,  face  mould  on  the  top  of  wreath,  with  the  points 
/and  TFover  the  points  S  and  0  at  the  ends  of  wreath,  then  mark 
it  underside  with  the  points  g  and  V,  over  the  point  H  and  D  at 
the  ends  of  wreath.  Proceed  to  work  the  wreath  as  before  de- 
scribed, to  .the  bevels  at  the  ends,  and  take  as  much  of  the  top  and 
bottom  sides  of  the  wreath  at  L  a  as  will  leave  the  rail  the  finished 
thickness. 


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